The smallest 5-digit number exactly divisible by 41 is
a) 1004
b) 10004
c) 10045
d) 10025

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Hint: In this question we need to find the greatest 4 digit number divisible by 41. Thus, by adding 41 to it will get the smallest 5 digit number divisible by 41.

Complete step-by-step answer:
We will try to find the greatest 4 digit number divisible by 41 first. For it we will use the trick of remainder formula which states that
$\text{Dividend=Quotient}\times \text{Divisor + Remainder }......\text{(i)}$
Here the remainder is less than the divisor. Dividend is the number which is being divided. Divisor is the number by which it is being divided and quotient is the maximum integer which can be multiplied to the divisor and the resultant would be still less than the dividend. Remainder is the difference between the dividend and the product of quotient and divisor.
In fact, this also means that if we subtract the remainder from the dividend, the resulting number would be exactly divisible by the divisor (as quotient is an integer), i.e.
  & \text{Dividend-Remainder=Quotient}\times \text{Divisor }......\text{(ii)} \\
 & \Rightarrow \dfrac{\text{Dividend-Remainder}}{\text{Divisor}}=\text{Quotient} \\
When we divide 10000 by 41, we get the quotient as 243 and 37 as the remainder. This means that 10000-37=9963 is the greatest 4 digit number divisible by 41.
Then, the smallest 5-digit number divisible by 41 would be 9963+41=10004 which matches option (b).
Hence, (b) is the correct answer.

Note: We can also write the whole method in terms of a formula:
The smallest n digit number divisible by an integer x =$({{10}^{n}}-\text{remainder obtained by dividing 1}{{\text{0}}^{n}}\text{by x)+x}$
So, using this formula we would have obtained,
The smallest 5 digit number divisible by 41=(10000-37)+41 = 10004 which matches the result obtained previously.